GATE Overflow Test Series | Discrete Mathematics | Test 3 | Question: 22

Question

Let $G$ be a cyclic group of order $96$, then the number of subgroups of $G$ is ________

Answer 1

A cyclic group of order $n$ has $d(n)$ number of subgroups, where,

$d(n) = $ number of positive divisors of $n.$

Divisors of $96 = 1,2,3,4,6,8,12,16,24,32,48,96$

$\therefore d(96) = 12$.

So, the correct answer is $12$.

Comments

Adding to the above answer, to find the number of divisors

1. Find prime factors of 96

           $96$ = $32 * 3$

                = $2^{5} * 3^{1}$ 

2. No of divisors of number n can be calculated as 

          $(a+1)(b+1)(c+1)…$ where a,b,c,… represent the power of each prime factor of n

           In case of $96$, we have $a=5$ and $b=1$ as $5$ and $1$ are powers of prime factors $2$ and $3$ respectively

           So , no of divisors of $96$ is $(a+1)*(b+1)$ = $6 * 2$ = $12$